Hello!
First, we need to calculate the moles of HNO₂ which are:
[tex]15,52 g HNO_2* \frac{1 mol HNO_2}{47,013 g HNO_2} =0,33 moles HNO_2[/tex]
The pKa for HNO₂ is pKa= -log(Ka)= -log(4,0 * 10⁻⁴)=3,39
Now, we can use the Henderson-Hasselbach equation to determine the moles of NaNO₂. Since both compounds (Weak base and acid) have the same volume, we can use moles instead of concentrations:
[tex]pH=pKa + log ( \frac{moles NaNO_2}{moles HNO_2} ) \\ \\ moles NaNO_2= moles HNO_2*10^{pH-pKa} \\ \\ moles NaNO_2= 0,33 moles* 10^{3,56-3,39}=0,49 moles NaNO_2 [/tex]
Finally, we go from moles of NaNO₂ to grams of NaNO₂ using the molar mass
[tex]0,49 moles NaNO_2* \frac{68,9953 g NaNO_2}{1 mol NaNO_2}=33,8077 g NaNO_2 [/tex]
So, the required amount of NaNO₂ is 33,8077 g
Have a nice day!
